Chapter 2, Problem 2.1P

A column in a building is subjected to the following load effects:

9 kips compression from dead load

5 kips compression from roof live load

6 kips compression from snow

7 kips compression from 3 inches of rain accumulated on the roof

8 kips compression from wind

a. If lead and resistance factor design is used, determine the factored load (required strength) to be used in the design of the column. Which AISC load combination controls?

b. What is the required design strength of the Column?

c. What is the required nominal strength of the column for a resistance factor $\varphi$ of 0.90?

d. If allowable strength design is used, determine the required load capacity (required strength) to be used in the design of the column. Which AISC load combination controls?

e. What is the required nominal strength of the column for a safety factor $\Omega$ of 1.67?

To determine

(a)

The maximum factored load and the controlling AISC load combination by using load and resistance factor design.

Answer to Problem 2.1P

The maximum factored load is $26\text{kips}$.

The AISC combination that controls is $1.2D+1.6(L_{\text{r}}\text{or}S\text{or}R)+\left(0.5L\text{or}0.5W\right)$.

Explanation of Solution

Given data:

The compression dead load is $9\text{kips}$.

The compression live load is $0\text{kips}$.

The compression snow load is $6\text{kips}$.

The compression roof live load is $5\text{kips}$.

The compression wind load is $8\text{kips}$.

The compression rain accumulated on the roof from $3\text{inches}$ is $7\text{kips}$.

Concept used:

The load and resistance design for load combinations is given below:

1. Combination $1$: $1.4D$
2. Combination $2$: $1.2D+1.6L+0.5(L_{\text{r}}\text{or}S\text{or}R)$
3. Combination $3$: $1.2D+1.6(L_{\text{r}}\text{or}S\text{or}R)+\left(0.5L\text{or}0.5W\right)$
4. Combination $4$: $1.2D+1.0W+0.5(L_{\text{r}}\text{or}S\text{or}R)$
5. Combination $5$: $0.9D+1.0W$

Here, the dead load is $D$, the live load is $L$, the roof live load is $L_{\text{r}}$, the snow load is $S$, the rain or ice load is $R$ and the wind load is $W$.

Determining the maximum factored load, depending upon the $5$ load combinations.

Write the expression for load combination $1$.

$L_1=1.4D$ ...... (I)

Here, load combination $1$ is $L_1$.

Substitute $9\text{kips}$ for $D$ in Equation (I).

$\begin{array}{c}{L}_1=1.4\times 9\text{kips}\\ =12.6\text{kips}\end{array}$

Write the expression for load combination $2$.

$L_2=1.2D+1.6L+0.5(L_{\text{r}}\text{or}S\text{or}R)$ ...... (II)

Here, load combination $2$ is $L_2$.

Substitute $9\text{kips}$ for $D$, $0\text{kips}$ for $L$ and $7\text{kips}$ for $(L_{\text{r}}\text{or}S\text{or}R)$ in Equation (II).

$\begin{array}{c}{L}_2=\left(1.2\times 9\text{kips}\right)+\left(1.6\times 0\text{kips}\right)+\left(0.5\times 7\text{kips}\right)\\ =14.3\text{kips}\end{array}$

Write the expression for load combination $3$.

$L_3=1.2D+1.6(L_{\text{r}}\text{or}S\text{or}R)+\left(0.5L\text{or}0.5W\right)$ ...... (III)

Here, load combination $3$ is $L_3$.

Substitute $9\text{kips}$ for $D$, $0\text{kips}$ for $L$, $8\text{kips}$ for $W$ and $7\text{kips}$ for $(L_{\text{r}}\text{or}S\text{or}R)$ in Equation (III).

$\begin{array}{c}{L}_3=\left(1.2\times 9\text{kips}\right)+\left(1.6\times 7\text{kips}\right)+\left(0.5\times 8\text{kips}\right)\\ =26\text{kips}\end{array}$

Write the expression for load combination $4$.

$L_4=1.2D+1.0W+0.5(L_{\text{r}}\text{or}S\text{or}R)$ ...... (IV)

Here, load combination $4$ is $L_4$.

Substitute $9\text{kips}$ for $D$, $8\text{kips}$ for $W$, $0\text{kips}$ for $L$ and $0\text{kips}$ for $(L_{\text{r}}\text{or}S\text{or}R)$ in Equation (IV).

$\begin{array}{c}{L}_4=\left(1.2\times 9\text{kips}\right)+\left(1.0\times 8\text{kips}\right)+\left(0.5\times 7\text{kips}\right)\\ =22.3\text{kips}\end{array}$

Write the expression for load combination $5$.

$L_5=0.9D+1.0W$ ...... (V)

Here, load combination $5$ is $L_5$.

Substitute $9\text{kips}$ for $D$ and $8\text{kips}$ for $W$ in Equation (V).

$\begin{array}{c}{L}_5=\left(0.9\times 9\text{kips}\right)+\left(1.0\times 8\text{kips}\right)\\ =16.1\text{kips}\end{array}$

From the above $5$ load combinations $L_3$ is maximum.

Conclusion:

Thus, the maximum factored load and AISC combination is estimated bythis expression: $L_1=1.4D$

To determine

(b)

The required design strength of the column.

Answer to Problem 2.1P

The required design strength of the column is $26\text{kips}$.

Explanation of Solution

The load combination $L_3$ is the required design strength of the column b, because the load combination $L_3$ includes all the criteria like rain, wind, live load and dead load.

Conclusion:

Thus, the design strength of the column is estimated by the load combination and concept.

To determine

(c)

The required nominal strength for a resistance factor of $0.90$.

Answer to Problem 2.1P

The required nominal strength of the column for a resistance factor is $28.9\text{kips}$.

Explanation of Solution

Concept used:

Write the expression for nominal strength of the column for LRFD.

$R_{\text{n}}=\frac{R_{\text{u}}}{\varphi }$ ...... (VI)

Here, the maximum factored load is $R_{\text{u}}$, the nominal strength of the column is $R_{\text{n}}$ and the resistance factor is $\varphi$.

Substitute $0.90$ for $\varphi$ and $26\text{kips}$ for $R_{\text{u}}$ in Equation (VI).

$\begin{array}{c}{R}_{\text{n}}=\frac{26\text{kips}}{0.90}\\ =28.9\text{kips}\end{array}$

Conclusion:

Thus, the required nominal strength of the column for a resistance factor is calculated by the expression for nominal strength of LFRD.

To determine

(d)

The maximum load and the controlling AISC load combination by using allowable strength design (ASD).

Answer to Problem 2.1P

The maximum load is $17.85\text{kips}$.

The AISC combination that controls is $D+0.75L+0.75\left(0.6W\right)+0.75(L_{\text{r}}\text{or}S\text{or}R)$.

Explanation of Solution

Concept used:

The following are the load combination for ASD:

1. Combination $1$: $D$
2. Combination $2$: $D+L$
3. Combination $3$: $D+(L_{\text{r}}\text{or}S\text{or}R)$
4. Combination $4$: $D+0.75L+0.75(L_{\text{r}}\text{or}S\text{or}R)$
5. Combination $5$: $D+0.6W$
6. Combination $6$: $D+0.75L+0.75\left(0.6W\right)+0.75(L_{\text{r}}\text{or}S\text{or}R)$
7. Combination $5$: $0.6D+0.6W$

Write the expression for load combination $1$.

$L_1=D$ ...... (VII)

Here, load combination $1$ is $L_1$.

Substitute $9\text{kips}$ for $D$ in Equation (VII).

$L_1=9\text{kips}$

Write the expression for load combination $2$.

$L_2=D+L$ ...... (VIII)

Here, load combination $2$ is $L_2$.

Substitute $9\text{kips}$ for $D$ and $0\text{kips}$ for $L$ in Equation (VIII).

$\begin{array}{c}{L}_2=9\text{kips}+0\text{kips}\\ =9\text{kips}\end{array}$

Write the expression for load combination $3$.

$L_3=D+(L_{\text{r}}\text{or}S\text{or}R)$ ...... (IX)

Here, load combination $3$ is $L_3$.

Substitute $9\text{kips}$ for $D$ and $7\text{kips}$ for $(L_{\text{r}}\text{or}S\text{or}R)$ in Equation (IX).

$\begin{array}{c}{L}_3=9\text{kips}+7\text{kips}\\ =16\text{kips}\end{array}$

Write the expression for load combination $4$.

$L_4=D+0.75L+0.75(L_{\text{r}}\text{or}S\text{or}R)$ ...... (X)

Here, load combination $4$ is $L_4$.

Substitute $9\text{kips}$ for $D$, $0\text{kips}$ for $L$ and $7\text{kips}$ for $(L_{\text{r}}\text{or}S\text{or}R)$ in Equation (X).

$\begin{array}{c}{L}_4=9\text{kips}+\left(0.75\times 0\text{kips}\right)+\left(0.75\times 7\text{kips}\right)\\ =14.25\text{kips}\end{array}$

Write the expression for load combination $5$.

$L_5=D+0.6W$ ...... (XI)

Here, load combination $5$ is $L_5$.

Substitute $9\text{kips}$ for $D$ and $8\text{kips}$ for $W$ in Equation (XI).

$\begin{array}{c}{L}_5=9\text{kips}+\left(0.6\times 8\text{kips}\right)\\ =13.8\text{kips}\end{array}$

Write the expression for load combination $6$.

$L_6=D+0.75L+0.75\left(0.6W\right)+0.75(L_{\text{r}}\text{or}S\text{or}R)$ ...... (XII)

Here, load combination $6$ is $L_6$.

Substitute $9\text{kips}$ for $D$, $0\text{kips}$ for $L$, $8\text{kips}$ for $W$ and $7\text{kips}$ for $(L_{\text{r}}\text{or}S\text{or}R)$ in Equation (XII).

$\begin{array}{c}{L}_6=9\text{kips}+\left(0.75\times 0\text{kips}\right)+0.75\left(0.6\times 8\text{kips}\right)+\left(0.75\times 7\text{kips}\right)\\ =17.85\text{kips}\end{array}$

Write the expression for load combination $7$.

$L_7=0.6D+0.6W$ ...... (XIII)

Here, load combination $7$ is $L_7$.

Substitute $9\text{kips}$ for $D$ and $8\text{kips}$ for $W$ in Equation (XIII).

$\begin{array}{c}{L}_7=\left(0.6\times 9\text{kips}\right)+\left(0.6\times 8\text{kips}\right)\\ =10.2\text{kips}\end{array}$

From the above $7$ load combinations $L_6$ is maximum.

Conclusion:

Thus, the maximum load and AISC combination is controlled by the load combination and the expression L1= D

To determine

(e)

The required nominal strength of the column for a safety factor of $1.67$.

Answer to Problem 2.1P

The required nominal strength of the column for a safety factor is $29.80\text{kips}$.

Explanation of Solution

Concept used:

Write the expression for nominal strength of the column for safety factor.

$\begin{array}{l}{R}_{\text{a}}=\frac{R_{\text{n}}}{\Omega }\\ R_{\text{n}}=R_{\text{a}}\times \Omega \end{array}$ ...... (XIV)

Here, the required strength is $R_{\text{a}}$, the nominal strength of the column is $R_{\text{n}}$ and the safety factor is $\Omega$.

Substitute $1.67$ for $\Omega$ and $17.85\text{kips}$ for $R_{\text{a}}$ in Equation (XIV).

$\begin{array}{c}{R}_{\text{n}}=\left(17.85\text{kips}\right)\times \left(1.67\right)\\ =29.80\text{kips}\end{array}$

Conclusion:

Thus, the required nominal strength of the column for safety factor is calculated by expression for the nominal strength of the column for safety factor.